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findMinInRotatedSortedArrayII.java
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38 lines (38 loc) · 1.58 KB
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import java.util.Arrays;
// 这道题目在面试中不会让写完整的程序
// 只需要知道最坏情况下 [1,1,1....,1] 里有一个0
// 这种情况使得时间复杂度必须是 O(n)
// 因此写一个for循环就好了。
// 如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。
// 反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。
public class findMinInRotatedSortedArrayII {
public static void main(String[] args){
findMinInRotatedSortedArrayII obj = new findMinInRotatedSortedArrayII();
int[] nums = {2,2,2,2,2,2,2,2,3,4,0,1,2,2};
System.out.println(obj.findMin(nums));
}
private int findMin(int[] nums){
int left = 0;
int right = nums.length - 1;
int mid;
while(left+1 < right){
mid = left + (right - left)/2;
System.out.println(Arrays.toString(new int[]{left,mid,right}));
//System.out.println(Arrays.toString(new int[]{nums[left],nums[mid],nums[right]}));
if(nums[mid] > nums[right]){
left = mid;
}else if(nums[mid] < nums[right]){
right = mid;
}else{
// if mid equals to end, that means it's fine to remove end
// the smallest element won't be removed
right --;
}
}
if(nums[left] < nums[right]){
return nums[left];
}else{
return nums[right];
}
}
}